5 Marks Questions

Question 15 Marks

What is meant by stability of a coordination compound in solution? State the factors which govern stability of complexes.

Answer

The stability of a complex in a solution refers to the degree of association between the two species involved in a state of equilibrium. Stability can be expressed quantitatively in terms of stability constant or formation constant. $\text{M}+3\text{L}\leftarrow\rightarrow\text{ML}_{3}$ Stability constant, $\beta=\frac{[\text{ML}_{3}]}{[\text{M}][\text{L}]^{3}}$ For this reaction, the greater the value of the stability constant, the greater is the proportion of $\ce{ML^3}$ in the solution. Stability can be of two types:

Thermodynamic stability:

The extent to which the complex will be formed or will be transformed into another species at the point of equilibrium is determined by thermodynamic stability.

Kinetic stability:

This helps in determining the speed with which the transformation will occur to attain the state of equilibrium.
Factors that affect the stability of a complex are:

Charge on the central metal ion: Thegreater the charge on the central metal ion, the greater is the stability of the complex.
Basic nature of the ligand: A more basic ligand will form a more stable complex.
Presence of chelate rings: Chelation increases the stability of complexes.

View full question & answer→

MCQ 25 Marks

Match the complex ions given in Column $I$ with the hybridisation and number of unpaired electrons given in Column $II$ and assign the correct code:

	Column $I ($Complex ion$)$		Column $II ($Hybridisation, number of unpaired electrons$)$
$a.$	$\ce{[Cr(H_2O)_6]^{3+}}$	$1.$	$dsp^2, 1$
$b.$	$\ce{[Co(CN)_4]^{2-}}$	$2.$	$sp^3d^2, 5$
$c.$	$\ce{[Ni(NH_3)_6]^{2+}}$	$3.$	$d^2sp^3, 3$
$d.$	$\ce{[MnF_6]^{4-}}$	$4.$	$sp^3, 4$
		$5.$	$sp^3d^2, 2$

A
$\ce{A (3), B (1), C (5), D (2)}.$
✓
$\ce{A (4), B (3), C (2), D (1)}.$
C
$\ce{A (3), B (2), C (4), D (1)}.$
D
$\ce{A (4), B (1), C (2), D (3)}.$

Answer

Correct option: B.

$\ce{A (4), B (3), C (2), D (1)}.$

View full question & answer→

Question 35 Marks

Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory. $\ce{[CoF_6]^{3-}, [Fe(CN)_6]^{4-}}$ and $\ce{[Cu(NH_3)_6]^{2+}}$

Answer

According to spectrochemical series related ligands the given complexes can be arranged in a series in the order of increasing field strength as$\ce{- CN^- > NH_{3 }> F^-}$
Thus $CN^-$ and $NH_3$ are strong field ligand pair up the $t_{2g}$ electrons before filling $e_g$ set.
$[\text{CoF}_6]^{3-},\text{Co}^{3+}(\text{d}^6)\text{t}^4_{2\text{g}}\text{e}^2_\text{g},$

$[\text{Fe}(\text{CN})_6]^{4-},\text{Fe}^{2+}(\text{d}^6)\text{t}^6_{2\text{g}}\text{e}^3_\text{g},$

$[\text{Cu}(\text{NH}_3)_6]^{2+},\ \text{Cu}^{2+}(\text{d}^9)\text{t}^6_{2\text{g}}\text{e}^3_\text{g},$

View full question & answer→

MCQ 45 Marks

Match the complex ions given in Column $I$ with the colours given in Column $II$ and assign the correct code:

	Column $I ($Complex ion$)$		Column $II ($Colour$)$
$a.$	$[Co(NH_3)_6]^{3+}$	$1.$	Violet
$b.$	$[Ti(H_2O)_6]^{3+}$	$2.$	Green
$c.$	$[Ni(H_2O)_6]^{2+}$	$3.$	Pale blue
$d.$	$[Ni(H_2O)_4(en)]^{2+} (aq)$	$4.$	Yellowish orange
		$5.$	Blue

Code:

A
$\text{A (1), B (2), C (4), D (5).}$
B
$\text{A (4), B (3), C (2), D (1).}$
C
$\text{A (3), B (2), C (4), D (1).}$
✓
$\text{A (4), B (1), C (2), D (3).}$

Answer

Correct option: D.

$\text{A (4), B (1), C (2), D (3).}$

	Column $I ($Complex ion$)$		Column $II ($Colour$)$
$a.$	$[Co(NH_3)_6]^{3+}$	$4.$	Yellowish orange
$b.$	$[Ti(H_2O)_6]^{3+}$	$1.$	Violet
$c.$	$[Ni(H_2O)_6]^{2+}$	$2.$	Green
$d.$	$[Ni(H_2O)_4(en)]^{2+} (aq)$	$3.$	Pale blue

View full question & answer→

MCQ 55 Marks

Match the coordination compounds given in Column $I$ with the central metal atoms given in Column $II$ and assign the correct code:

	Column $I ($Coordination Compound$)$		Column $II ($Central metal atom$)$
$a.$	Chlorophyll	$1.$	Rhodium
$b.$	Blood pigment	$2.$	Cobalt
$c.$	Wilkinson catalyst	$3.$	Calcium
$d.$	Vitamin $B_{12}$	$4.$	Iron
		$5.$	Magnesium

Code:

✓
$\text{A (5), B (4), C (1), D (2).}$
B
$\text{A (3), B (4), C (5), D (1).}$
C
$\text{A (4), B (3), C (2), D (1).}$
D
$\text{A (3), B (4), C (1), D (2).}$

Answer

Correct option: A.

$\text{A (5), B (4), C (1), D (2).}$

	Column $I ($Coordination Compound$)$		Column $II ($Central metal atom$)$
$a.$	Chlorophyll	$5.$	Magnesium
$b.$	Blood pigment	$4.$	Iron
$c.$	Wilkinson catalyst	$1.$	Rhodium
$d.$	Vitamin $B_{12}$	$2.$	Cobalt

View full question & answer→

Question 65 Marks

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

$[\text{CoF}_6]^{3-}, [\text{Co(H}_2\text{O})_6]^{2+}, [\text{Co(CN)}_6]^{3-}$
$[\text{FeF}_6]^{3-}, [\text{Fe(H}_2\text{O})_6]^{2+}, [\text{Fe(CN})_6]^{4-}$

Answer

$\text{Co}^{2+}=3\text{d}^7$
Number of unpaired electrons = 4
Magnetic moment
$=\sqrt{\text{n}(\text{n}+2)}=\sqrt{4(4+2)}=4.9\text{B.M.}$
$[\text{Co(H}_2\text{O})_6]^{2+}:$

$\text{Co}^{2+}=3\text{d}^7$
Number of unpaired electrons = 3
Magnetic moment = $\sqrt{3(3+2)}=3.87\text{B.M.}$
$[\text{Co(CN)}_6]^{3-}:$

$\text{Co}^{3+}=3\text{d}^6$
Number of unpaired electrons = 0
Diamagnetic.

$\text{FeF}^{3-}_6:$

$\text{Fe}^{3+}=3\text{d}^5$
Number of unpaired electrons = 5
Magnetic moment = $\sqrt{5(5+2})=5.92\text{B.M.}$
$[\text{Fe(H}_2\text{O})_6]^{2+}:$

$\text{Fe}^{2+}=3\text{d}^6$
Number of unpaired electrons = 4
Magnetic moment = $\sqrt{4(4+2})=4.9\text{B.M.}$
$[\text{Fe(CN)}_6]^{4-}:$

$\text{Fe}^{2+}=3\text{d}^6$
Diamagnetic.

View full question & answer→

MCQ 75 Marks

Match the compounds given in Column $I$ with the oxidation state of cobalt present in it $($given in Column $II)$ and assign the correct code.

	Column $I ($Compound$)$		Column $II ($Oxidation state of $Co)$
$a.$	$\ce{[Co(NCS)(NH_3)_5](SO_3)}$	$1.$	$+4$
$b.$	$\ce{[Co(NH_3)_4Cl_2]SO_4}$	$2.$	$0$
$c.$	$\ce{Na_4[Co(S_2O_3)_3]}$	$3.$	$+1$
$d.$	$\ce{[Co_2(CO)_8]}$	$4.$	$+2$
		$5.$	$+3$

Code:

A
$\ce{A (1), B (2), C (4), D (5).}$
B
$\ce{A (4), B (3), C (2), D (1).}$
✓
$\ce{A (5), B (1), C (4), D (2).}$
D
$\ce{A (4), B (1), C (2), D (3).}$

Answer

Correct option: C.

$\ce{A (5), B (1), C (4), D (2).}$

	$Column $I ($Compound$)$		Column $II ($Oxidation state of $Co)$
$a.$	$[Co(NCS)(NH_3)_5](SO_3)$	$5.$	$+3$
$b.$	$[Co(NH_3)_4Cl_2]SO_4$	$1.$	$+4$
$c.$	$Na_4[Co(S_2O_3)_3]$	$4.$	$+2$
$d.$	$[Co_2(CO)_8]$	$2.$	$0$

Explanation:
Oxdiation state of $CMI ($central metal ion$)$ can be calculated by considering the oxidation state of whole molecule is equal to charge present on coordination sphere.

$[Co(NCS)(NH_3)_5]SO_3$

Let oxidation state of $Co$ be $ x.$
$x - 1 + × 0 = +2$
$x = +2 + 1 = +3$

$[Co(NH_3)_4Cl_2]SO_4$

Let oxidation state of $Co = x$
$\Rightarrow x + 4 \times 0 + 2 \times (-1) = +2$
$\Rightarrow x - 2 = +2$
$x = 4$

$Na_4[Co(S_2O_3)_3]$

Let oxidation state of $Co = x$
$x + 3 × (-2) = -4$
$x - 6 = -4$
$x = -4 + 6 = +2$

$[Co(CO)_8]$

Let oxidation state of $Co = x$
$x - 8 × 0 = 0$
$x = 0$
Hence, correct choice is $(c).$

View full question & answer→

MCQ 85 Marks

Match the complex species given in Column $I$ with the possible isomerism given in Column $II$ and assign the correct code:

	Column $I ($Complex species$)$		Column $II ($Isomerism$)$
$a.$	$[Co(NH_3)_4C_{l2}]^+$	$1.$	Optical
$b.$	$cis-[Co(en)_2Cl_2]^+$	$2.$	Ionisation
$c.$	$[Co(NH_3)_5 (NO_2)]Cl_2$	$3.$	Coordination
$d.$	$[Co(NH_3)_6][Cr(CN)_6]$	$4.$	Geometrical
		$5.$	Linkage

Code:

A
$\text{A (1), B (2), C (4), D (5).}$
B
$\text{A (4), B (3), C (2), D (1).}$
C
$\text{A (4), B (1), C (5), D (3).}$
D
$\text{A (4), B (1), C (2), D (3).}$

Answer

	Column $I ($Complex species$)$		Column $II ($Isomerism$)$
$a.$	$[Co(NH_3)_4C_{l2}]^+$	$4.$	Geometrical
$b.$	$cis-[Co(en)_2Cl_2]^+$	$1.$	Optical
$c.$	$[Co(NH_3)_5 (NO_2)]Cl_2$	$2.$	Ionisation
$d.$	$[Co(NH_3)_6][Cr(CN)_6]$	$3.$	Coordination

Explanation:
Isomerism in coordination compound is decided by type of ligands and geometry of coordination and arrangement of ligands.

$[Co(NH_3)_4Cl_2]+$ shows geometrical isomerism due to presence of two types of ligand whose $[Co(NH_3)_4Cl_2]+$ arrangement around central metal ion.

cis$-[Co(en)_2Cl_2]+$ shows optical isomer due to its non$-$superimposable mirror image relationship.

$[Co(NH_3)_5(NO_2)]Cl_2$ shows ionization isomer due to its interchanging ligand from outside the ionization sphere.
$[Co(NH_3)_6][Cr(CN)_6]$ shows coordination isomer due to interchanging of ligand in between two metal ions from one coordination sphere to another coordination sphere.

View full question & answer→

Question 95 Marks

$\ce{COSO_4Cl.5NH_3}$ exists in two isomeric forms $‘A’$ and $‘B’$. Isomer $‘A’$ reacts with $\ce{AgNO_3}$ to give white precipitate, but does not react with $\ce{BaCl_2.}$ Isomer $‘B’$ gives white precipitate with $\ce{BaCl_2} $ but does not react with $\ce{AgNO_3}$. Answer the following questions.

Identify $‘A’$ and $‘B’$ and write their structural formulas.
Name the type of isomerism involved.
Give the $\text{IUPAC}$ name of $‘A’$ and $‘B’.$

Answer

$\ce{CoSO_4Cl.5NH_3:}$

Isomer A reacts with $\ce{AgN0_3}$ but not with $\ce{BaCl_2},$ it shows it has $CP$ ion outside the coordination sphere.

Hence, $\ce{A = [Co(NH_3)_5SO_4]Cl}$
Isomer B reacts with $BaCl_2$ but not with $\ce{AgNO_3 }$, it shows it has $\ce{SO_4^-}$ outside the coordination sphere.
Hence, $\ce{B = [CO(NH_3)5Cl]S0_4}$

Ionisation isomerism.
$A =$ Pentaamminesulphatocobalt $(III)$ chloride and.

$B =$ Pentaamminesulphatocobalt $(III)$ sulphate.

View full question & answer→

Chemistry 5 Marks Questions

Take a timed test